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    can someone help me with June 2006 C3 Q8 b, I understand the multiplying out the brackets but I don't understand where sin comes from in the mark scheme


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    (Original post by Delusional09)
    Stuck on this, any help is appreciated! Stuck on the whole question!
    a) what do you get if you do cos(2x+x)?
    b)i) multiply cosx/1+sinx by cosx and then multiply 1+sinx/cosx by 1+sinx and go on from here.
    ii) 2secx=4
    secx=2
    now what is cosx? and then sketch the graph to find other values.
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    (Original post by Delusional09)
    Stuck on this, any help is appreciated! Stuck on the whole question!
    which paper is this? just so i can check if i the got right answer
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    (Original post by justinawe)
    Actually, after subbing in values, I think they may actually be equivalent. But I'm not sure how to justify it.

    I'm not sure if you can just square root both sides like you did though, you need to account for negative roots as well.
    So if I left my answer as \dfrac{16x^2}{(x^2-1)^2}=\dfrac{(y^2-1)^2}{y^2}, would that probably be okay?
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    Anyone doing Solomon/Elmwood papers?

    I had to get an A* in maths or further maths (aced C4 in January) and they helped a lot for C3 (I didn't do any C4 Solomon papers because it was Christmas).
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    (Original post by VSEPR)
    Anyone doing Solomon/Elmwood papers?

    I had to get an A* in maths or further maths (aced C4 in January) and they helped a lot.
    i've done all C3 soloman papers, should i then do the elmwood papers for C3? or do actually past papers, cuz exam is this thursday? thanks
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    (Original post by masryboy94)
    i've done all C3 soloman papers, should i then do the elmwood papers for C3? or do actually past papers, cuz exam is this thursday? thanks
    I'd suggest reading Examiner's Reports as they tell you where most people slip up.

    Other than that do all the peripheral stuff (good night's sleep, breakfast, water) as those final few marks can be achieved on the day by being psyched.
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    (Original post by Lamalam)
    hey guys, I am quite confused about the edexcel c3 june 2006 q5a, Attachment 225233 and here is how I worl the questionAttachment 225239, am I correct?

    i am confused about the cancelling of cos^2 (2K), why can it be eliminated?
    my teacher told me that
    if x^2 = 2x , I cannot cross out anything but putting the 2x on the same side as x^2 and carry on working.
    why in the q5, the cos^2 (2k) can be eliminated? will this cause a loss of answer?
    in what situation can a quantity be eliminated?

    thank you soooo much!! please help!
    Erm.. well.. you're multiplying through by cos^2 2k , so where you've got your 5th line, you multiply through by cos^2 2k , and now you'll have 2(2k-1) + 2sin2kcos2k=0.. so on your 6th line you shouldn't have cos^2 2k at the bottom. and where you thought you cancelled off, you're meant to use sin2A=2sinAcosA formula. So 2sin2kcos2k = sin4k... and that's how you get the sin4k from the equation. everything else you did is correct. Hope this helped
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    (Original post by masryboy94)
    which paper is this? just so i can check if i the got right answer
    its Jan 2008 q6. Thanks
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    (Original post by brittanna)
    For part (a), use the identity in the question and the fact that cos3x=cos(2x+x).

    For part (b), try expressing the two fractions as a single fraction.
    thanks
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    (Original post by brittanna)
    So if I left my answer as \dfrac{16x^2}{(x^2-1)^2}=\dfrac{(y^2-1)^2}{y^2}, would that probably be okay?
    I think that should be okay.
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    (Original post by Delusional09)
    Erm.. well.. you're multiplying through by cos^2 2k , so where you've got your 5th line, you multiply through by cos^2 2k , and now you'll have 2(2k-1) + 2sin2kcos2k=0.. so on your 6th line you shouldn't have cos^2 2k at the bottom. and where you thought you cancelled off, you're meant to use sin2A=2sinAcosA formula. So 2sin2kcos2k = sin4k... and that's how you get the sin4k from the equation. everything else you did is correct. Hope this helped
    on my 5th line, i am actually putting two fractions into one, since the bottom of the middle fraction is cos2k, i multiply it bu another cos2k in order for it to be combined into one fraction (which i think it is a normal practice), am i wrong for doing this ? can you explain ??

    thank you !!!!!
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    Anyone know if proof by contradiction will ever actually come up?
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    C4- can someone help me with b and c? Please please I would appreciate it so much!

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    (Original post by Lamalam)
    on my 5th line, i am actually putting two fractions into one, since the bottom of the middle fraction is cos2k, i multiply it bu another cos2k in order for it to be combined into one fraction (which i think it is a normal practice), am i wrong for doing this ? can you explain ??

    thank you !!!!!
    Yeah, I totally get where you got it from. What your doing is right.. its just another way of doing it. So the last line.. if you're wondering about the cancelling out that you were talking about.. you don't need to cancel it out.. just take it to the other side to multiply by 0 and you have your answer. Which leads me to a question.. does anybody have any tips to remembering the Trig Identities and formulae?
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    (Original post by ChelseaSam)
    Anyone know if proof by contradiction will ever actually come up?
    no it's not on the spec.
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    (Original post by yaboy)
    I know how to draw it but Im useto using the cast diagram and that seems to be pretty useless atm.
    The CAST diagram is fine.
    If you have one solution for tan the next one is pi radians on.
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    C4: How do you integrate cot2x? Formula book states that the integral of cotx is ln|sinx| so would it be... 0.5ln|sin2x|?
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    (Original post by ChelseaSam)
    C4: How do you integrate cot2x? Formula book states that the integral of cotx is ln|sinx| so would it be... 0.5ln|sin2x|?
    Try using the substituion let u = 2x

    Spoiler:
    Show

    ∫ cot(2x) dx

    Let:
    u = 2x
    du = 2 dx
    (1/2)du = dx

    Remember this identity: cotθ = cosθ/sinθ
    = ∫ cot(u) * (1/2)du
    = 1/2 ∫ cot(u) du
    = 1/2 ∫ cos(u)/sin(u) du

    Let:
    w = sin(u)
    dw = cos(u) du

    = 1/2 ∫ dw / w
    = 1/2 ∫ 1/w dw
    = 1/2ln|w| + C
    = 1/2ln|sin(u)| + C
    = 1/2ln|sin(2x)| + C

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    (Original post by Boy_wonder_95)
    Try using the substituion let u = 2x

    Spoiler:
    Show

    ∫ cot(2x) dx

    Let:
    u = 2x
    du = 2 dx
    (1/2)du = dx

    Remember this identity: cotθ = cosθ/sinθ
    = ∫ cot(u) * (1/2)du
    = 1/2 ∫ cot(u) du
    = 1/2 ∫ cos(u)/sin(u) du

    Let:
    w = sin(u)
    dw = cos(u) du

    = 1/2 ∫ dw / w
    = 1/2 ∫ 1/w dw
    = 1/2ln|w| + C
    = 1/2ln|sin(u)| + C
    = 1/2ln|sin(2x)| + C

    Ahhh ok makes sense when you integrate from first principles I was just wondering if reverse chain rule applied to those given integrals. Thanks
 
 
 
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